The following problem was featured as a challenge in a previous exam paper and has left me stumped.

Compute the following limit: $$ \lim_{n \to \infty} \frac{1}{n^{2013}} \sum_{k=1}^n k^{2012} $$

Any help or hints to get me on the right track would be greatly appreciated!

  • 2
    $\begingroup$ Hint: Riemann sum $\endgroup$
    – zhw.
    May 24 '16 at 19:24
  • $\begingroup$ If I'm not mistaken L'Hôpital's Rule also works, reducing to the right answer. $\endgroup$
    – KR136
    May 24 '16 at 20:17
  • $\begingroup$ L'Hôpital's Rule works for functions $\endgroup$
    – user340508
    May 24 '16 at 20:37

Two approaches.

(I) $$ 2012!\binom{k}{2012}\leq k^{2012} \leq 2012!\binom{k+2012}{2012}, $$ hence: $$ 2012!\binom{n+1}{2013}\leq \sum_{k=1}^{n}k^{2012} \leq 2012!\binom{n+2013}{2013} $$ and if we divide both sides by $n^{2013}$ and take the limit as $n\to +\infty$ we get that the limit is $\color{red}{\large\frac{1}{2013}}$ by squeezing.

(II) By Riemann sums, $$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^{2012} = \int_{0}^{1}x^{2012}\,dx = \color{red}{\frac{1}{2013}}.$$



This is kind of complicated, so the first thing you want to do is find a pattern somehow. Let's say the general form of the limit is the following: $$\lim_{n \to \infty} \frac 1 {n^{m+1}} \sum_{k=0}^n k^m$$

The limit you want to find is $m=2012$, but start with $m=2,3,4,5,...$ first to find a pattern. If you need help finding a formula for the $\sum_{k=0}^n k^m$ part, here are the formulas for this sum from $m=1$ to $m=10$.


The Stolz–Cesaro theorem is one of most standard methods for such examples.



or with proof:


For $b_n=n^{2013}$ and $a_n=\sum_{k=1}^{n}k^{2012}$ we have the case when it works.

$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{(n+1)^{2012}}{(n+1)^{2013}-n^{2013}}=\frac{(n+1)^{2012}}{(n+1)^{2012}+(n+1)^{2011}n^{1}+\dots}$$ divide by $(n+1)^{2012}$ to find the limit: $\frac{1}{2013}$.

Of course you can generalize for other exponents instead of 2013, we have year 2016 currently :-]


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